Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{6k(5k - 6)}{5k} \div \dfrac{40k - 48}{7} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{6k(5k - 6)}{5k} \times \dfrac{7}{40k - 48} $ When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 6k(5k - 6) \times 7 } { 5k \times (40k - 48) } $ $ y = \dfrac {7 \times 6k(5k - 6)} {5k \times 8(5k - 6)} $ $ y = \dfrac{42k(5k - 6)}{40k(5k - 6)} $ We can cancel the $5k - 6$ so long as $5k - 6 \neq 0$ Therefore $k \neq \dfrac{6}{5}$ $y = \dfrac{42k \cancel{(5k - 6})}{40k \cancel{(5k - 6)}} = \dfrac{42k}{40k} = \dfrac{21}{20} $